Derivatives · Definition

Implicit differentiation

calculusderivatives

Before you start you need to know

Some curves, such as the circle, are not graphs of a single function of \(x\). Implicit differentiation finds \(\frac{dy}{dx}\) without solving for \(y\).

Definition

When \(y\) is defined implicitly by an equation in \(x\) and \(y\), both sides are differentiated with respect to \(x\), remembering that \(y\) depends on \(x\) (chain rule), and then \(\frac{dy}{dx}\) is solved for.

\[x^{2}+y^{2}=25\ \Rightarrow\ 2x+2y\,y'=0\ \Rightarrow\ y'=-\frac{x}{y}\]

Sometimes \(y\) is not solved for: for example the circle \(x^2+y^2=25\). Still, near a point \(y\) is a function of \(x\), and we can differentiate both sides with respect to \(x\) treating \(y\) as a function of \(x\).

Every time you differentiate a term containing \(y\), the chain rule adds a factor \(y'\) (or \(dy/dx\)). At the end, solve for \(y'\).

It is the tool for finding tangents of curves that are not graphs of a single function, and for deriving general logs and roots.

Think about it

Find \(y'\) for \(x^2+y^2=25\).

\(2x+2yy'=0\), so \(y'=-x/y\).

Exercises 10

What comes next

References

  • James Stewart, Calculus: Early Transcendentals, 8.ª edición, Cengage Learning, 2015.